Find the absolute minimum and absolute maximum values of f on the given interval. f (x) = ln (x2 + 3x + 4), [-2, 2]

The vertex (minimum) of the quadratic ax² + bx + c is located at x=-b / (2a). This means the minimum value of f (x) will be found at x = - 3 / (2*1) = - 1.5.

Since the vertex of the quadratic is less than 0, the maximum value of the quadratic will be found at x=2, the end of the interval farthest from the vertex.

On the given interval, ...

the absolute minimum value of f is f (-1.5) = ln (1.75) ≈ 0.559616

the absolute maximum value of f is f (2) = ln (14) ≈ 2.639057